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Daily NEET MCQ- March 31, 2022 - NEET MCQ


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10 Questions MCQ Test - Daily NEET MCQ- March 31, 2022

Daily NEET MCQ- March 31, 2022 for NEET 2024 is part of NEET preparation. The Daily NEET MCQ- March 31, 2022 questions and answers have been prepared according to the NEET exam syllabus.The Daily NEET MCQ- March 31, 2022 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Daily NEET MCQ- March 31, 2022 below.
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Daily NEET MCQ- March 31, 2022 - Question 1

Aristotle classified the organisms based on their _________

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 1

Aristotle classified the organisms based on their morphological characteristics. His classification neglected characteristics like nature of the cell wall, evolutionary relationships and method of reproduction.

Daily NEET MCQ- March 31, 2022 - Question 2

Aristotle’s classification contained ________

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 2

Aristotle’s classification contained two kingdoms i.e. Plants and Animals. Plants are sub-divided into herbs, shrubs and tree. Animals are further sub-divided into red blooded and non- red blooded animals.

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Daily NEET MCQ- March 31, 2022 - Question 3

Linnaeus classified organisms into ________

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 3

Aristotle classified the organisms into Plants and Animals whereas Carolus Linnaeus classified the organisms into Plantae and Animalia. Whittaker’s 5 kingdom classification contained Monera, Protista, Fungi, Plantae and Animalia.

Daily NEET MCQ- March 31, 2022 - Question 4

What is the time period of kinetic energy of a particle in SHM,if the time period of SHM is 4s?

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 4

The kinetic energy of a particle in SHM is given by:
0.5mA2w2cos2(wt + a), where a is the phase constant.
cos2(wt + a) = (1 + cos2(wt+a))/2.
Therefore the time period of oscillation of kinetic energy is 4/2 = 2s.

Daily NEET MCQ- March 31, 2022 - Question 5

Consider a particle undergoing an SHM of amplitude A & angular frequency w. What is the magnitude of displacement from the mean position when kinetic energy is equal to the magnitude of potential energy?

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 5

Let x = Asin(wt).
∴ Kinetic energy = 1/2 mA2w2cos2(wt)
potential energy = 1/2kA2sin2(wt).
According to given condition:
1/2mA2w2cos2(wt) = 1/2A2w2sin2(wt)
using, k = mw2, we get:
tan2(wt) = 1.
∴ wt = π/4.
∴ x = Asin(π/4) = A/√2.

Daily NEET MCQ- March 31, 2022 - Question 6

A particle is in SHM. When displacement is ‘a‘ potential energy is 8J. When displacement is ‘a‘ potential energy is 18J. What will be the potential energy when displacement is ‘a+b’?

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 6

1/2ka2 = 8 & 1/2kb2 = 18.
We have to find value of 1/2k(a + b) 2.
∴ 1/2k(a + b)2 = 8 + 18 + kab
= 26+(k*4*6/k)
= 26 + 24 = 50J.

Daily NEET MCQ- March 31, 2022 - Question 7

If the potential energy of the particle is given by: U = x- 8x + 16. It starts from rest from x = 0. What will be its maximum speed during SHM? Assume the mass of the particle to be 1kg.

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 7

F = -dU/dx = 8 – 2x.
Particle starts from x=0, so force will increase its speed till force becomes zero at x=4.
Let acceleration be ‘a’.
a = dv/dt = v dv/dx.
And a = F/m = F/1 = 8-2x.
adx = vdv. Now, integrate both sides. The limits of dx will be 0 to 4 and the limits of dv will be 0 to v.
(8x – x2)04 = v2/2.
∴ v = 4√2 m/s.

Daily NEET MCQ- March 31, 2022 - Question 8

2-chloropropane and 1-chloropropane exhibit ____________ isomerism.

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 8

When two or more compounds have the same molecular formula but the different position of functional groups are substituents, they are called positional isomers and the phenomenon is called position isomerism. Here 2-chloropropane and 1-chloropropane differ in position so they exhibit position isomerism.

Daily NEET MCQ- March 31, 2022 - Question 9

How many planes of symmetry does a meso compound have?

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 9

The compound whose half part of a molecule is a mirror image of the other half, is called meso form. Generally, a meso compound has two or more chiral centers and one plane of symmetry. the compound in meso form is optically inactive due to internal compensation.

Daily NEET MCQ- March 31, 2022 - Question 10

If a compound has 3 chiral carbons What is the number of optically active isomers?

Detailed Solution for Daily NEET MCQ- March 31, 2022 - Question 10

The number of optically active isomers for a compound is given as 2n where, n represents the number of chiral carbons in that particular carbon, here as the compound has 3 chiral carbons n = 3, so the number of optically active isomers equals 23 = 8.

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